﻿ 一种新非Lipschitz条件下倒向随机微分方程的L2解 L2 Solutions of BSDEs with a New Kind of Non-Lipschitz

Vol. 08  No. 08 ( 2019 ), Article ID: 31615 , 6 pages
10.12677/AAM.2019.88155

L2 Solutions of BSDEs with a New Kind of Non-Lipschitz

Shiyu Li, Liping Dan, Lufan Yang

Faculty of Science, Jiangxi University of Sciences and Technology, Ganzhou Jiangxi

Received: Jul. 15th, 2019; accepted: Aug. 1st, 2019; published: Aug. 8th, 2019

ABSTRACT

The classical backward stochastic differential equation (BSDE) is driven by the Brownian motion, but Brownian motion is a very special stochastic process, so the application of backward stochastic differential equation is quite limited. In this paper, we are interested in solving one-dimensional backward stochastic differential equations (BSDEs) with a new kind of non-Lipschitz coefficients [1] . We establish an existence and uniqueness result of solutions in L2.

Keywords:Backward Stochastic Differential Equation, Continuous Local Martingale, Non-Lipschitz, Existence, Uniqueness

1. 引言

2. 主要结果

$\left(\Omega ,,F,，{\left\{{F}_{t}\right\}}_{t\ge 0},P\right)$ 为一个带信息流的完备的概率空间，其中流 ${\left\{{F}_{t}\right\}}_{t\ge 0}$ 满足通常条件，记 $\mathcal{P}$ 为可料 σ 域。 $M=\left\{{M}_{t},{F}_{t}:0\le t<\infty \right\}$ 为一个连续局部鞅，并且 ${M}_{0}=0$$〈M〉$ 为M的平方变差过程。 $T>0$ 为一个任意固定的数，称为时间区间。

1) 用 ${L}_{M}\left(0,T;{R}^{n}\right)$ 表示所有使得 ${‖x‖}_{M}^{2}=E{\int }_{0}^{T}{|y\left(s\right)|}^{2}d{〈M〉}_{s}<+\infty$${F}_{t}$ -适应的 ${R}^{n}$ 值的过程 $x=x\left(s\right)$ 的集合。当 $n=1$ 时简记为 ${L}_{M}$

2) 用 ${{L}^{\prime }}_{M}\left(0,T;{R}^{n×d}\right)$ 表示所有使得 ${‖y‖}_{M}^{2}=E{\int }_{0}^{T}{|z\left(s\right)|}^{2}d{〈M〉}_{s}<+\infty$${F}_{t}$ 可料的 ${R}^{n×d}$ 值的过程 $y=y\left(s\right)$ 的集合。当 $n=d=1$ 时简记为 ${{L}^{\prime }}_{M}$

3) 用 ${L}^{2}\left(\Omega ,{F}_{T},P;{R}^{n}\right)$ 表示所有满足${F}_{T}$ -可测的 ${R}^{n}$ 值的随机变量 $\xi$ 的集合。当 $n=1$ 时简记为 ${L}^{2}\left(\Omega ,{F}_{T},P\right)$

(1)

(H1) $g\left(\cdot ,0,0\right)\in {L}_{M}$

(H2) 存在一个单调不减凹函数 $\rho \left(\cdot \right):{R}^{+}\to {R}^{+}$，使得 $\forall {y}_{1},{y}_{2}\in R,z\in R$ , $\text{d}P×\text{d}t-a.e.$

${|g\left(w,t,{y}_{1},z\right)-g\left(w,t,{y}_{2},z\right)|}^{2}\le \rho \left({|{y}_{1}-{y}_{2}|}^{2}\right)$

(H3)存在一个常数 $C\ge 0$，使得 $\forall y\in R,{z}_{1},{z}_{2}\in R$ , $\text{d}P×\text{d}t-a.e.$

${|g\left(w,t,y,{z}_{1}\right)-g\left(w,t,y,{z}_{2}\right)|}^{2}\le C\left({|{z}_{1}-{z}_{2}|}^{2}\right)$

(H4) $E\left[{\left({\int }_{0}^{T}|g\left(t,0,0\right)|\text{d}{〈M〉}_{t}\right)}^{2}\right]<+\infty$

3. 引理

${y}_{t}^{0}=0;\text{\hspace{0.17em}}{y}_{t}^{n}=\xi +{\int }_{t}^{T}\left[g\left(s,{y}_{s}^{n-1},{z}_{s}^{n}\right)\right]\text{d}{〈M〉}_{s}-{\int }_{t}^{T}{z}_{s}^{n}\text{d}{M}_{s},0\le t\le T$(2)

$|g\left(s,{y}_{s}^{n-1},0\right)|\le |g\left(s,0,0\right)|+{\rho }^{\frac{1}{2}}\left({|{y}_{s}^{n-1}|}^{2}\right)\le |g\left(s,0,0\right)|+{A}^{\frac{1}{2}}\left(|{y}_{s}^{n-1}|+1\right)$

$E\left[{\left({\int }_{0}^{T}|g\left(s,{y}_{s}^{n-1},0\right)|d{〈M〉}_{s}\right)}^{2}\right]\le 4E\left[{\left({\int }_{0}^{T}|g\left(s,0,0\right)|d{〈M〉}_{s}\right)}^{2}\right]+A{\left(2T\right)}^{2}\left(E\left[\underset{s\in \left[0,T\right]}{\mathrm{sup}}{|{y}_{s}^{n-1}|}^{2}\right]+1\right)$

$E\left[\underset{s\in \left[t,T\right]}{\mathrm{sup}}{|{y}_{s}^{n+m}-{y}_{s}^{n}|}^{2}\right]\le \frac{1}{2}{\text{e}}^{{c}_{1}\left(T-t\right)}{\int }_{t}^{T}\rho \left(E\left[{|{y}_{s}^{n+m-1}-{y}_{s}^{n-1}|}^{2}\right]\right)\text{d}{〈M〉}_{s}$ (3)

$E\left[\underset{s\in \left[t,T\right]}{\mathrm{sup}}|{z}_{s}^{n+m}-{z}_{s}^{n}|\right]\le K\left\{E\left[\underset{s\in \left[t,T\right]}{\mathrm{sup}}{|{y}_{s}^{n+m}-{y}_{s}^{n}|}^{2}\right]+{\int }_{t}^{T}\rho \left(E\left[{|{y}_{s}^{n+m-1}-{y}_{s}^{n-1}|}^{2}\right]\right)d{〈M〉}_{s}\right\}$ (4)

${y}_{t}={\int }_{t}^{T}\left[{f}_{n,m}\left(s,{z}_{s}\right)\right]d{〈M〉}_{s}-{\int }_{t}^{T}{z}_{s}\text{d}{M}_{s},0\le t\le T$ (5)

$|{f}_{n,m}\left(s,{z}_{s}\right)|\le {\rho }^{\frac{1}{2}}\left({|{y}_{s}^{n+m-1}-{y}_{s}^{n-1}|}^{2}\right)+C|z|$ (6)

(6)式意味着方程(5)的生成元 ${f}_{n,m}\left(s,{z}_{s}\right)$ 满足文献 [1] 命题1中的假设(A)，即 $\psi \left(\cdot \right)\equiv 0$$\lambda =C$${f}_{t}\equiv 0$${\phi }_{t}={\rho }^{\frac{1}{2}}\left({|{y}_{s}^{n+m-1}-{y}_{s}^{n-1}|}^{2}\right)$。又因为 $\rho \left(\cdot \right)$ 是一个凹函数，所以由文献 [1] 命题1和命题2，应用Fubini定理和

Jensen不等式，即可得(3)式和(4)式。 证毕。

$|g\left(s,{y}_{s}^{n-1},z\right)|\le |g\left(s,{y}_{s}^{n-1},z\right)-g\left(s,0,0\right)|+|g\left(s,0,0\right)|\le {\rho }^{\frac{1}{2}}\left({|{y}_{s}^{n-1}|}^{2}\right)+C|z|+|g\left(s,0,0\right)|$

$E\left[\underset{r\in \left[t,T\right]}{\mathrm{sup}}{|{y}_{r}^{n}|}^{2}\right]\le {\mu }_{t}+\frac{1}{2}{\text{e}}^{{c}_{3}\left(T-t\right)}{\int }_{t}^{T}\rho \left(E\left[{|{y}_{s}^{n-1}|}^{2}\right]\right)d{〈M〉}_{s}$ (7)

$N=2{\mu }_{0}+2AT$${T}_{1}=\mathrm{max}\left\{T-\frac{\mathrm{ln}2}{{c}_{1}},T-\frac{\mathrm{ln}2}{{c}_{3}},T-\frac{1}{2}A,0\right\}$，其中 ${c}_{1}$ 是引理1中的，A是注1中的,则对，有

$\frac{1}{2}{\text{e}}^{{c}_{1}\left(T-t\right)}\le 1,\frac{1}{2}{\text{e}}^{{c}_{3}\left(T-t\right)},A\left(T-t\right)\le \frac{1}{2}$ (8)

$E\left[\underset{r\in \left[t,T\right]}{\mathrm{sup}}{|{y}_{r}^{n}|}^{2}\right]\le {\mu }_{0}+\frac{1}{2}{\text{e}}^{{c}_{3}\left(T-t\right)}{\int }_{t}^{T}\rho \left(E\left[{|{y}_{s}^{n-1}|}^{2}\right]\right)d{〈M〉}_{s},t\in \left[{T}_{1},T\right]$ (9)

$E\left[\underset{r\in \left[t,T\right]}{\mathrm{sup}}{|{y}_{r}^{n}|}^{2}\right]\le N$，证毕。

4. 定理1的证明

${\phi }_{0}\left(t\right)=\underset{t}{\overset{T}{\int }}\rho \left(N\right)\text{d}{〈M〉}_{s};\text{\hspace{0.17em}}{\phi }_{n+1}\left(t\right)=\underset{t}{\overset{T}{\int }}\rho \left({\phi }_{n}\left(s\right)\right)\text{d}{〈M〉}_{s}$ (10)

$\forall t\in \left[{T}_{1},T\right]$，由引理2，得

${\phi }_{0}\left(t\right)=\underset{t}{\overset{T}{\int }}\rho \left(N\right)\text{d}{〈M〉}_{s}\le M$

${\phi }_{1}\left(t\right)=\underset{t}{\overset{T}{\int }}\rho \left({\phi }_{0}\left(s\right)\right)d{〈M〉}_{s}\le \underset{t}{\overset{T}{\int }}\rho \left(N\right)d{〈M〉}_{s}={\phi }_{0}\left(t\right)\le M$

${\phi }_{2}\left(t\right)=\underset{t}{\overset{T}{\int }}\rho \left({\phi }_{1}\left(s\right)\right)d{〈M〉}_{s}\le \underset{t}{\overset{T}{\int }}\rho \left({\phi }_{0}\left(s\right)\right)d{〈M〉}_{s}={\phi }_{1}\left(t\right)\le M$

$0\le {\phi }_{n+1}\left(t\right)\le {\phi }_{n}\left(t\right)\le \cdots \le {\phi }_{1}\left(t\right)\le {\phi }_{0}\left(t\right)\le M$

$E\left[\underset{r\in \left[t,T\right]}{\mathrm{sup}}{|{y}_{r}^{n}|}^{2}\right]\le N$

$E\left[\underset{r\in \left[t,T\right]}{\mathrm{sup}}{|{y}_{r}^{1+m}-{y}_{r}^{1}|}^{2}\right]\le {\int }_{t}^{T}\rho \left(E\left[{|{y}_{s}^{m}|}^{2}\right]\right)d{〈M〉}_{s}\le \underset{t}{\overset{T}{\int }}\rho \left(N\right)d{〈M〉}_{s}={\phi }_{0}\left(t\right)\le M$

$E\left[\underset{r\in \left[t,T\right]}{\mathrm{sup}}{|{y}_{r}^{2+m}-{y}_{r}^{2}|}^{2}\right]\le {\int }_{t}^{T}\rho \left(E\left[{|{y}_{r}^{\text{1+}m}-{y}_{r}^{1}|}^{2}\right]\right)d{〈M〉}_{s}\le \underset{t}{\overset{T}{\int }}\rho \left({\phi }_{0}\left(s\right)\right)d{〈M〉}_{s}={\phi }_{1}\left(t\right)\le M$

$E\left[\underset{r\in \left[{T}_{1},T\right]}{\mathrm{sup}}{|{y}_{r}^{\text{n+}m}-{y}_{r}^{n}|}^{2}\right]\le {\phi }_{n-1}\left({T}_{1}\right)\to 0,n\to \infty$

${\left\{{y}_{t}^{n}\right\}}_{n\ge 1}$ 是cauchy序列,又因为 $\rho \left(\cdot \right)$ 是一个连续函数，由引理1中的(4)式知， ${\left\{{z}_{t}^{n}\right\}}_{n\ge 1}$ 也是cauchy序列，它们的极限分别记为 ${\left\{{y}_{t}\right\}}_{t\in \left[{T}_{1},T\right]}$${\left\{{y}_{t}\right\}}_{t\in \left[{T}_{1},T\right]}$。令 $n\to \infty$，对(2)式取极限，可得 $\left\{{y}_{t},{z}_{t}\right\}$ 是具有参数 $\left(\xi ,T,g\right)$ 的BSDE在 $\left[{T}_{1},T\right]$${L}^{2}$ 解。

${y}_{t}={\int }_{t}^{T}\left[\stackrel{^}{g}\left(s,{y}_{s},{z}_{s}\right)\right]d{〈M〉}_{s}-{\int }_{t}^{T}{z}_{s}\text{d}{M}_{s},0\le t\le T$ (11)

1中的假设(A)。

$E\left[{|{y}_{t}^{1}-{y}_{t}^{2}|}^{2}\right]\le \frac{1}{2}{\text{e}}^{{c}_{4}\left(T-t\right)}{\int }_{t}^{T}\rho \left(E\left[{|{y}_{s}^{1}-{y}_{s}^{2}|}^{2}\right]\right)d{〈M〉}_{s}$ (12)

$E\left[\left({\int }_{t}^{T}{|{z}_{s}^{1}-{z}_{s}^{2}|}^{2}\text{d}{〈M〉}_{s}\right)\right]\le {c}_{5}\left\{E\left[\underset{s\in \left[t,T\right]}{\mathrm{sup}}{|{y}_{s}^{1}-{y}_{s}^{2}|}^{2}\right]+\rho \left(E\left[\underset{s\in \left[t,T\right]}{\mathrm{sup}}{|{y}_{s}^{1}-{y}_{s}^{2}|}^{2}\right]\right)\right\}$(13)

L2 Solutions of BSDEs with a New Kind of Non-Lipschitz[J]. 应用数学进展, 2019, 08(08): 1321-1326. https://doi.org/10.12677/AAM.2019.88155

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